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Rationalize the difference in boiling points for each of the following pairs of substances: a. \(n\) -pentane \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(36.2^{\circ} \mathrm{C}\) b. \(\mathrm{HF} \quad 20^{\circ} \mathrm{C}\) \(\mathrm{HCl} \quad-85^{\circ} \mathrm{C}\) c. \(\mathrm{HCl} \quad-85^{\circ} \mathrm{C}\) \(\mathrm{LiCl} \quad 1360^{\circ} \mathrm{C}\) d. \(n\) -pentane \(\quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) \(36.2^{\circ} \mathrm{C}\) \(n\) -hexane \(\quad \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3} \quad 69^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The differences in boiling points for the given pairs of substances can be explained by different factors, such as the types of intermolecular forces present, molecular mass, and molecular size. In the case of n-pentane and n-hexane, London dispersion forces are the primary factor, and the higher molecular mass of n-hexane leads to a higher boiling point. For HF and HCl, the stronger dipole-dipole interactions in HF, due to the higher electronegativity of Fluorine, result in a higher boiling point for HF. Lastly, for HCl and LiCl, the stronger ionic bonds in LiCl cause it to have a significantly higher boiling point compared to HCl, which has weaker dipole-dipole interactions.

Step by step solution

01

Understanding n-pentane and n-hexane

Both n-pentane and n-hexane are linear hydrocarbons with respective chemical formulas: n-pentane: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (boiling point: \(36.2^{\circ} \mathrm{C}\)) n-hexane: \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\) (boiling point: \(69^{\circ} \mathrm{C}\)) The primary type of intermolecular force present in these substances is London dispersion forces, which are caused by temporary fluctuations in the electron density of the molecules.
02

Comparing the molecular mass

n-pentane has 5 carbon atoms, while n-hexane has 6 carbon atoms. The molecular mass of n-hexane is greater than n-pentane due to the extra carbon atom. Since London dispersion forces increase with molecular mass, this leads to a greater boiling point for n-hexane.
03

Understanding HF and HCl

Both HF and HCl are hydrogen halides with the following boiling points: HF: \(20^{\circ} \mathrm{C}\) HCl: \(-85^{\circ} \mathrm{C}\) The primary intermolecular forces present in these substances are dipole-dipole interactions, due to the electronegativity difference between hydrogen and the halogen atom.
04

Comparing electronegativity and boiling points

Fluorine is more electronegative than Chlorine, which results in a stronger dipole-dipole interaction in an HF molecule. This enhanced interaction results in a higher boiling point for HF as compared to HCl, even though the molecular mass of HCl is greater than that of HF.
05

Understanding HCl and LiCl

HCl: \(-85^{\circ} \mathrm{C}\) LiCl: \(1360^{\circ} \mathrm{C}\) HCl is a polar covalent compound, while LiCl is an ionic compound. The primary intermolecular force in HCl is the dipole-dipole interaction, whereas the forces between LiCl molecules are ionic bonds.
06

Comparing ionic and covalent bonds

Ionic bonds are significantly stronger than dipole-dipole interactions, leading to a much higher boiling point for LiCl compared to HCl. In conclusion, the difference in boiling points between the given pairs of substances can be rationalized by understanding the intermolecular forces, molecular mass, and size of the molecules and their implications on boiling points.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Boiling Point
The boiling point of a substance is the temperature at which it transforms from a liquid to a gas. It is an essential property related to the substance's physical characteristics and intermolecular forces. Stronger intermolecular forces mean more energy is required to separate the molecules, which results in a higher boiling point. For example, as seen in the comparison between n-pentane and n-hexane, n-hexane has a higher boiling point because it possesses more atoms, thus exhibiting stronger intermolecular forces. This makes it more energy-demanding to convert from liquid to gas.
In short, boiling points can vary widely across different substances due to differences in molecular size and the types and strengths of intermolecular forces at play.
London Dispersion Forces
London dispersion forces are the weakest type of intermolecular force and are present in all molecules, regardless of whether they are polar or nonpolar. They are caused by the temporary fluctuations in electron density that create a transient dipole, inducing a dipole in neighboring molecules. These forces become more significant as the size of the molecule increases, due to the larger electron clouds that can polarize more easily.
For instance, n-pentane and n-hexane mainly rely on London dispersion forces. n-hexane, with its larger molecular size compared to n-pentane, exhibits stronger dispersion forces, resulting in a higher boiling point. Despite being weak individually, London dispersion forces can collectively have a substantial overall impact on the physical properties of a substance.
Dipole-Dipole Interactions
Dipole-dipole interactions occur between molecules that have permanent dipoles, which arise due to differences in electronegativity between atoms in a molecule. These interactions are stronger than London dispersion forces and significantly affect boiling points. The strength of dipole-dipole interactions is influenced by how polar the molecules are; the greater the difference in electronegativity, the stronger the interaction.
For example, HF and HCl exhibit dipole-dipole interactions. HF has a notably higher boiling point than HCl because fluorine is more electronegative than chlorine, resulting in stronger dipole-dipole interactions in HF. Enhanced interaction and the accompanying hydrogen bonding in HF require more energy for the molecules to separate, thus raising the boiling point compared to HCl.
Ionic Bonds
Ionic bonds are strong electrostatic forces present between positively and negatively charged ions. These bonds are much stronger than intermolecular forces like dipole-dipole interactions and London dispersion forces. As a result, substances with ionic bonds, such as LiCl, have extraordinarily high boiling points compared to covalent compounds.
In the exercise, LiCl's boiling point is significantly higher than that of HCl. This difference arises because LiCl contains ionic bonds, which are markedly stronger and require substantially more energy to overcome. The strength of ionic bonds stems from the strong attraction between ions, resulting in very high melting and boiling points, further illustrating the profound impact of bonding types on physical properties.

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Most popular questions from this chapter

Amino acids are the building blocks of the body's worker molecules called proteins. When two amino acids bond together, they do so through the formation of a peptide linkage, and a dipeptide is formed. Consider the following tripeptide formed when three alanine amino acids bond together: What types of interparticle forces could be present in a sample of this tripeptide?

An ice cube tray contains enough water at \(22.0{ }^{\circ} \mathrm{C}\) to make 18 ice cubes that each have a mass of \(30.0 \mathrm{~g} .\) The tray is placed in a freezer that uses \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) as a refrigerant. The heat of vaporization of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) is \(158 \mathrm{~J} / \mathrm{g} .\) What mass of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) must be vaporized in the refrigeration cycle to convert all the water at \(22.0^{\circ} \mathrm{C}\) to ice at \(-5.0^{\circ} \mathrm{C}\) ? The heat capacities for \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(I)\) are \(2.03 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), respectively, and the enthalpy of fusion for ice is \(6.02 \mathrm{~kJ} / \mathrm{mol}\).

Consider the following enthalpy changes: $$ \begin{aligned} \mathrm{F}^{-}+\mathrm{HF} \longrightarrow \mathrm{FHF}^{-} \quad \Delta H=-155 \mathrm{~kJ} / \mathrm{mol} \\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}+\mathrm{HF} \longrightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{O}-\mathrm{-HF} \\ \Delta H=-46 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{HOH}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}--\mathrm{HOH}(\text { in ice }) \\ \Delta H=-21 \mathrm{~kJ} / \mathrm{mol} \end{aligned} $$ How do the strengths of hydrogen bonds vary with the electronegativity of the element to which hydrogen is bonded? Where in the preceding series would you expect hydrogen bonds of the following type to fall?

What is an alloy? Explain the differences in structure between substitutional and interstitial alloys. Give an example of each type.

The unit cell of MgO is shown below. Does \(\mathrm{MgO}\) have a structure like that of \(\mathrm{NaCl}\) or \(\mathrm{ZnS} ?\) If the density of \(\mathrm{MgO}\) is \(3.58 \mathrm{~g} / \mathrm{cm}^{3}\), estimate the radius (in centimeters) of the \(\mathrm{O}^{2-}\) anions and the \(\mathrm{Mg}^{2+}\) cations.

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